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Math Help - True or false analysis.

  1. #1
    Super Member Showcase_22's Avatar
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    True or false analysis.

    Which of the following are true and which false? Give proofs or counterexamples.
    i). If x \rightarrow f(x)^2 (this is a mapping, I couldn't quite find the right arrow) is continous at c then f is differentiable at c.
    I said this was false and my counterexample was f(x)=| \sqrt{x}|.

    This function is continous at 0 but f(x)^2=|x| which is not differentiable at 0.

    ii). If a,b) \rightarrow \mathbb{R}" alt="\exists \ ga,b) \rightarrow \mathbb{R}" /> and a,b) \rightarrow \mathbb{R}" alt="Za,b) \rightarrow \mathbb{R}" /> such that f(x)=g(x)Z(x) then f is differentiable at c if:

    a). Z is bounded
    b). g is differentiable at c with g'(c)=0 and
    c). g(c)=0;
    I'm not really sure about this one.

    I think it's true but I don't know how to prove it.

    iii). In ii) f is differentiable assuming only (a) and (c).
    I put false and used the counterexample f(x)=x \sin \left( \frac{1}{x} \right) when c=0.

    Z(x)= \sin \left( \frac{1}{x} \right) and g(x)=x.

    In ii). f is differentiable assuming only (a) and (b).
    Once again I put false and used the counterexample f(x)=5 \sin \left( \frac{1}{x} \right) where c=0.

    Where g(x)=5 and Z(x)= \sin \left( \frac{1}{x} \right)

    In (ii). f is differentiable assuming only (b) and (c).
    This is false because Z(x) could be discontinous.

    My counterexample was f(x)=x^2. \begin{cases} -1 \ \ \ \ \ x \in{\mathbb{Q}}\\<br />
0 \ \ \ \ \ x \in \mathbb{R}-{\mathbb{Q}} \end{cases} at c=0.

    Here g(x)=x^2 and Z(x)=\begin{cases} -1 \ \ \ \ \ x \in{\mathbb{Q}}\\<br />
0 \ \ \ \ \ x \in \mathbb{R}-{\mathbb{Q}} \end{cases}.

    are these correct?
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    I said this was false and my counterexample was f(x)=| \sqrt{x}|.

    This function is continous at 0 but f(x)^2=|x| which is not differentiable at 0.
    You mean f(x)=\sqrt{|x|} I suppose ?
    If so it is OK
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  3. #3
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    Quote Originally Posted by Showcase_22 View Post
    I'm not really sure about this one.

    I think it's true but I don't know how to prove it.
    Yes it is true

    \frac{f(x) - f(c)}{x-c} = \frac{g(x)Z(x) - g(c)Z(c)}{x-c} = \frac{g(x)Z(x)}{x-c} = Z(x) \:\frac{g(x)}{x-c} = Z(x) \:\frac{g(x) - g(c)}{x-c}

    I think that you can finish.
    This will give you ideas for the other cases.
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post
    I put false and used the counterexample f(x)=x \sin \left( \frac{1}{x} \right) when c=0.

    Z(x)= \sin \left( \frac{1}{x} \right) and g(x)=x.
    I do not agree with your counter-example since g is not defined in 0 (even if its limit is 0).

    Take Z(x) = 1 \mbox{ and } g(x) = \sqrt{|x|} over (-1,1) for instance

    Z is bounded and g(0)=0 therefore (a) and (c) are verified

    f(x) = \sqrt{|x|} is not differentiable in 0
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